IB Diploma ProgrammeHL

Simple harmonic motion

Physics HL Chapter Notes

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Simple harmonic motionWave modelWave phenomenaStanding waves and resonanceDoppler effect
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In this chapter

  1. 1What is Simple Harmonic Motion?
  2. 2Visualising SHM: Graphs and Phase
  3. 3The Equations of SHM
  4. 4Energy Transformations in SHM
  5. 5Real-World Examples: Pendulums and Springs

1. What is Simple Harmonic Motion?

An oscillation is a repeating back-and-forth motion about a central point, called the equilibrium position. Simple Harmonic Motion (SHM) is a special, very common type of oscillation. For a motion to be considered SHM, it must meet two precise conditions: the acceleration of the object must be directly proportional to its displacement from the equilibrium position, and the acceleration must always be directed in the opposite direction to the displacement (i.e., always towards the equilibrium position). This can be summarised by the relationship a ∝ -x. The force that causes this acceleration is called the restoring force. Because F = ma, the restoring force is also proportional to the displacement and acts towards the equilibrium point (F ∝ -x).

a = -ω²x

F ∝ -x

T = 1/f

ω = 2πf = 2π/T

Key term

Restoring Force: A force that always acts to pull or push an oscillating system back towards its equilibrium position.

Examiner insight

Examiners frequently test the definition of SHM. A clear, two-part statement linking acceleration and displacement is essential for full marks.

Common pitfall

Forgetting to state that the acceleration (or force) is in the opposite direction to the displacement; just stating it is proportional is only half the definition.

Worked example 14 marks

The acceleration 'a' of a particle varies with its displacement 'x' from a fixed point as shown in the graph. Explain how the graph shows that the particle is undergoing SHM and determine the period of the oscillation. The graph is a straight line passing through the origin, (0.20 m, -5.0 m s⁻²) and (-0.20 m, 5.0 m s⁻²).

  1. 1

    Step 1: State the conditions for SHM. For SHM, acceleration 'a' must be directly proportional to displacement 'x' (a ∝ x), and in the opposite direction.

  2. 2

    Step 2: Relate the graph to the conditions. The graph is a straight line passing through the origin, which shows that a is directly proportional to x. The graph has a negative gradient (as x increases, a becomes more negative), which shows that the acceleration is always in the opposite direction to the displacement.

  3. 3

    Step 3: Use the defining equation of SHM, a = -ω²x. The gradient of the a-x graph is equal to -ω².

  4. 4

    Step 4: Calculate the gradient. Gradient = Δa / Δx = (-5.0 - 5.0) / (0.20 - (-0.20)) = -10.0 / 0.40 = -25 s⁻².

  5. 5

    Step 5: Equate the gradient to -ω² to find ω. -ω² = -25, so ω² = 25, and ω = 5.0 rad s⁻¹.

  6. 6

    Step 6: Calculate the period T using the formula T = 2π/ω. T = 2π / 5.0 ≈ 1.26 s.

Recap

  • Simple Harmonic Motion (SHM) is a periodic motion about an equilibrium position.
  • The defining condition for SHM is that acceleration is proportional to displacement and in the opposite direction (a ∝ -x).
  • The restoring force is the net force causing SHM, always directed towards equilibrium.
  • Amplitude (x₀) is the maximum displacement from the equilibrium position.
  • Period (T) is the time for one complete oscillation, while frequency (f) is the number of oscillations per second.
  • Angular frequency (ω) relates period and frequency by ω = 2π/T = 2πf.

Quick check

  1. State the two defining conditions for an object to be undergoing simple harmonic motion.2 marks

2. Visualising SHM: Graphs and Phase

The motion of an object in SHM can be represented by graphs of displacement, velocity, and acceleration against time. If the object starts at the equilibrium position (x=0) moving in the positive direction, the displacement-time (x-t) graph is a sine curve. The velocity is the gradient of the x-t graph, so the velocity-time (v-t) graph is a cosine curve. The acceleration is the gradient of the v-t graph, resulting in an a-t graph that is an inverted sine curve. This reveals key phase relationships: velocity leads displacement by 90° (π/2 radians), and acceleration leads velocity by 90° (π/2 radians). This means acceleration and displacement are 180° (π radians) out of phase, or in 'antiphase'—when one is at its positive maximum, the other is at its negative maximum.

Key term

Phase Difference: The fraction of an oscillation cycle by which one oscillating quantity leads or lags another, usually expressed in radians or degrees.

Fun fact

The motion of a shadow cast by an object moving in a uniform circle on a wall is a perfect physical demonstration of simple harmonic motion.

Worked example 14 marks

An object oscillates with SHM. The graph shows the variation of its displacement x with time t. From the graph, determine:(a) the amplitude,(b) the period,(c) the frequency, and(d) the angular frequency of the oscillation. The graph is a sine wave starting at (0,0), reaching a peak of 4.0 cm at t=0.5 s, and completing one full cycle at t=2.0 s.

  1. 1

    Step 1 (a): The amplitude is the maximum displacement from the equilibrium position. From the graph, the peak value of x is 4.0 cm. So, amplitude x₀ = 4.0 cm or 0.040 m.

  2. 2

    Step 2 (b): The period is the time taken for one complete oscillation. From the graph, the motion repeats every 2.0 seconds. So, period T = 2.0 s.

  3. 3

    Step 3 (c): The frequency is the reciprocal of the period, f = 1/T. f = 1 / 2.0 s = 0.50 Hz.

  4. 4

    Step 4 (d): The angular frequency is given by ω = 2πf or ω = 2π/T. Using T = 2.0 s, ω = 2π / 2.0 = π rad s⁻¹ ≈ 3.14 rad s⁻¹.

Worked example 22 marks

For the oscillation described in the previous example, at what time is the velocity first at its positive maximum, and at what time is the acceleration first at its negative maximum?

  1. 1

    Step 1: Recall the phase relationships. Velocity is maximum when displacement is zero and the gradient of the x-t graph is steepest and positive. Acceleration is maximum negative when displacement is maximum positive.

  2. 2

    Step 2: Analyse the velocity. The x-t graph is a sine curve. The gradient is steepest and positive at the start, t=0 s, and again at t=2.0 s. So the velocity is first at its positive maximum at t=0 s.

  3. 3

    Step 3: Analyse the acceleration. Acceleration is in antiphase with displacement (a ∝ -x). Displacement is at its positive maximum (x = +4.0 cm) for the first time at t = 0.5 s. Therefore, the acceleration is at its negative maximum at this same time, t = 0.5 s.

Recap

  • A displacement-time graph for SHM is sinusoidal (a sine or cosine curve).
  • The amplitude is the peak value on the displacement-time graph.
  • The period is the time for one full wave cycle on any of the graphs.
  • Velocity leads displacement by π/2 radians (90°).
  • Acceleration leads velocity by π/2 radians (90°).
  • Acceleration and displacement are in antiphase, meaning they are π radians (180°) out of phase.

Quick check

  1. If an object in SHM is at its maximum positive displacement, what are its instantaneous velocity and acceleration?2 marks

3. The Equations of SHM

The graphical description of SHM can be expressed mathematically. The displacement x at time t is given by x = x₀sin(ωt) if the oscillation starts at x=0, or x = x₀cos(ωt) if it starts at the maximum displacement x=x₀. Here, x₀ is the amplitude and ω is the angular frequency. The velocity v and acceleration a can be found from the displacement equation. For x = x₀sin(ωt), the velocity is v = ωx₀cos(ωt) and the acceleration is a = -ω²x₀sin(ωt). Notice that this last equation simplifies to a = -ω²x, the fundamental definition of SHM. A very useful equation that is independent of time links velocity and displacement: v = ±ω√(x₀² - x²). This allows you to calculate the speed of the oscillator at any position if you know the amplitude and angular frequency.

x = x₀sin(ωt) or x = x₀cos(ωt)

v = ωx₀cos(ωt) or v = -ωx₀sin(ωt)

a = -ω²x₀sin(ωt) or a = -ω²x₀cos(ωt)

v_max = ωx₀

a_max = ω²x₀

v = ±ω√(x₀² - x²)

Key term

Angular Frequency (ω): A measure of the rate of oscillation in radians per unit time, defined as ω = 2πf, and measured in radians per second (rad s⁻¹).

Common pitfall

Confusing frequency (f in Hz) with angular frequency (ω in rad s⁻¹). Always check which one the question gives or asks for, and remember the conversion ω = 2πf.

Worked example 15 marks

A particle's motion is described by the equation x = 0.05cos(10πt), where x is in meters and t is in seconds. Find:(a) the amplitude,(b) the frequency,(c) the maximum speed, and(d) the speed when the particle is at x = 0.03 m.

  1. 1

    Step 1 (a): Compare the given equation with the standard form x = x₀cos(ωt). By comparison, the amplitude x₀ = 0.05 m.

  2. 2

    Step 2 (b): By comparison, the angular frequency ω = 10π rad s⁻¹. Since ω = 2πf, we can find the frequency f. f = ω / 2π = 10π / 2π = 5.0 Hz.

  3. 3

    Step 3 (c): The maximum speed is given by v_max = ωx₀. v_max = (10π rad s⁻¹) × (0.05m) = 0.5π m s⁻¹ ≈ 1.57 m s⁻¹.

  4. 4

    Step 4 (d): Use the velocity-displacement equation v = ±ω√(x₀² - x²). v = ±10π × √(0.05² - 0.03²) = ±10π × √(0.0025 - 0.0009) = ±10π × √0.0016 = ±10π × 0.04 = ±0.4π m s⁻¹. The speed is the magnitude, so speed = 0.4π m s⁻¹ ≈ 1.26 m s⁻¹.

Recap

  • The displacement in SHM is described by x = x₀sin(ωt) or x = x₀cos(ωt).
  • The choice between sine and cosine depends on the starting position at t=0.
  • Maximum speed occurs at x=0 and is v_max = ωx₀.
  • Maximum acceleration occurs at x=±x₀ and is a_max = ω²x₀.
  • The equation v = ±ω√(x₀² - x²) allows you to find speed at any displacement.

Quick check

  1. An oscillator has an amplitude of 10 cm and a period of 4.0 s. What is its maximum acceleration?3 marks

4. Energy Transformations in SHM

In an ideal simple harmonic oscillator with no friction, the total mechanical energy (E_total) is conserved. This energy is continuously transformed between kinetic energy (KE) and potential energy (PE). At the extreme points of the oscillation (x = ±x₀), the object is momentarily stationary, so its KE is zero and its PE is maximum. As the object moves towards the equilibrium position (x=0), its speed increases, converting PE into KE. At the equilibrium position, the speed is maximum, so KE is maximum, and the PE is zero (by definition). The total energy at any point is E_total = KE + PE. This total energy is constant and is equal to the maximum PE or the maximum KE: E_total = ½kx₀² = ½m(v_max)². Graphically, a plot of energy vs. displacement shows the PE as a 'U'-shaped parabola, the KE as an 'n'-shaped parabola, and the total energy E_total as a constant horizontal line.

KE = ½mv²

PE = ½kx² = ½mω²x²

E_total = KE + PE

E_total = ½kx₀² = ½mω²x₀²

Key term

Total Energy (in SHM): The constant sum of the kinetic and potential energies of the oscillator, which is proportional to the square of the amplitude.

Examiner insight

Marks are often awarded for correctly sketching energy-displacement graphs. Ensure your PE graph is a parabola opening upwards from the origin, the KE graph is a parabola opening downwards from its maximum at the origin, and the total energy is a horizontal line.

Worked example 15 marks

A 0.50 kg mass is attached to a spring and oscillates with an amplitude of 10 cm and a period of 2.0 s. Calculate(a) the total energy of the system and(b) the displacement at which the kinetic energy is equal to the potential energy.

  1. 1

    Step 1: List the known values and convert to SI units. m = 0.50 kg, x₀ = 10 cm = 0.10 m, T = 2.0 s.

  2. 2

    Step 2: Calculate the angular frequency, ω. ω = 2π/T = 2π / 2.0 = π rad s⁻¹.

  3. 3

    Step 3 (a): Calculate the total energy using E_total = ½mω²x₀². E_total = ½ × 0.50 × (π)² × (0.10)² = 0.25 × π² × 0.01 ≈ 0.0247 J.

  4. 4

    Step 4 (b): Set KE = PE. We know that E_total = KE + PE. So, if KE = PE, then E_total = PE + PE = 2PE.

  5. 5

    Step 5: Substitute the formulas for E_total and PE. ½mω²x₀² = 2 × (½mω²x²).

  6. 6

    Step 6: Simplify and solve for x. The ½mω² terms cancel, leaving x₀² = 2x². So, x² = x₀²/2. x = ±x₀/√2.

  7. 7

    Step 7: Calculate the numerical value. x = ±0.10 / √2 ≈ ±0.071 m.

Recap

  • In SHM, energy converts between kinetic (KE) and potential (PE).
  • Total mechanical energy (KE + PE) is constant in the absence of damping.
  • KE is maximum at the equilibrium position (x=0); PE is maximum at the amplitudes (x=±x₀).
  • PE is zero at the equilibrium position; KE is zero at the amplitudes.
  • Total energy is proportional to the square of the amplitude (E_total ∝ x₀²).
  • KE and PE are equal when the displacement is x = ±x₀/√2.

Quick check

  1. If the amplitude of an SHM oscillation is doubled, by what factor does the total energy of the oscillator increase?1 mark

5. Real-World Examples: Pendulums and Springs

Two classic physical systems that exhibit SHM are the mass-spring system and the simple pendulum. For a mass 'm' on a spring of spring constant 'k', Hooke's Law (F = -kx) provides the restoring force. Since F=ma, we get ma = -kx, or a = -(k/m)x. This perfectly matches the SHM definition a = -ω²x, where ω² = k/m. This gives the period of a mass-spring system as T = 2π√(m/k). A simple pendulum consists of a point mass 'm' on a massless string of length 'L'. The restoring force is a component of gravity, F = -mgsinθ. For small angles of swing (typically < 10°), we can use the small angle approximation, sinθ ≈ θ (where θ is in radians). Since the arc length displacement is x = Lθ, the force becomes F ≈ -(mg/L)x. This again fits the SHM form, with ω² = g/L. Therefore, for small oscillations, the period of a simple pendulum is T = 2π√(L/g).

T = 2π√(m/k) (for a mass-spring system)

T = 2π√(L/g) (for a simple pendulum with small angles)

Key term

Small Angle Approximation: An approximation for small angles (typically < 10°) where sin(θ) ≈ θ (with θ in radians), which allows a simple pendulum's motion to be modelled as SHM.

Common pitfall

Applying the simple pendulum formula T = 2π√(L/g) to oscillations with large amplitudes, where the motion is periodic but not simple harmonic.

Fun fact

Early seismographs, used to detect earthquakes, were based on the principle of a large-mass pendulum. The frame of the instrument moves with the ground, but the inertia of the heavy pendulum bob keeps it almost stationary, allowing the relative motion to be recorded.

Worked example 14 marks

A 200 g mass is attached to a vertical spring, causing it to extend by 15 cm to a new equilibrium position. The mass is then pulled down a further 5 cm and released.(a) Calculate the spring constant k.(b) Calculate the period of the resulting oscillations.

  1. 1

    Step 1 (a): At the new equilibrium, the spring force balances the weight of the mass. F = mg = kΔL. Convert mass and extension to SI units: m = 0.200 kg, ΔL = 0.15 m.

  2. 2

    Step 2 (a): Rearrange to find k. k = mg / ΔL = (0.200 kg × 9.81 m s⁻²) / 0.15 m = 13.08 N m⁻¹ ≈ 13 N m⁻¹.

  3. 3

    Step 3 (b): The period of a mass-spring system is given by T = 2π√(m/k).

  4. 4

    Step 4 (b): Substitute the values. T = 2π√(0.200 kg / 13.08 N m⁻¹) = 2π√0.01529 ≈ 0.777 s. (Note: the extra 5 cm pull-down is the amplitude and does not affect the period).

Worked example 23 marks

A grandfather clock needs a pendulum with a period of 2.00 s (a 'seconds pendulum'). What length must the pendulum have? (Assume g = 9.81 m s⁻²).

  1. 1

    Step 1: Start with the formula for the period of a simple pendulum, T = 2π√(L/g).

  2. 2

    Step 2: Rearrange the formula to make L the subject. First, square both sides: T² = 4π²(L/g).

  3. 3

    Step 3: Isolate L: L = gT² / (4π²).

  4. 4

    Step 4: Substitute the given values. T = 2.00 s and g = 9.81 m s⁻². L = (9.81 × 2.00²) / (4π²) = (9.81 × 4) / (4π²) = 9.81 / π² ≈ 0.994 m.

Recap

  • A mass-spring system undergoes SHM with period T = 2π√(m/k).
  • The period of a mass-spring system depends on mass and spring constant, not amplitude.
  • A simple pendulum undergoes SHM only for small angles of oscillation.
  • The period of a simple pendulum is T = 2π√(L/g) for small angles.
  • The period of a simple pendulum depends on its length and the local gravity, not on its mass or amplitude (for small angles).

Quick check

  1. If you take a pendulum clock to the Moon, where the acceleration due to gravity is about 1/6th of that on Earth, will it run faster or slower?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. An object of mass 400 g undergoes SHM with a frequency of 2.5 Hz and an amplitude of 8.0 cm. Calculate (a) the maximum velocity and (b) the maximum kinetic energy of the object.5 marks
  2. A particle oscillates with SHM. Its velocity v is related to its displacement x by the equation v² = 144(0.06² - x²), where all quantities are in SI units. Determine the amplitude and the period of the oscillation.4 marks
  3. Sketch a graph to show the variation of potential energy (PE) with displacement (x) for a particle undergoing SHM. On the same axes, sketch the variation of kinetic energy (KE) with displacement. Label the points where x=0 and x=±x₀.3 marks
  4. A simple pendulum has a period of 1.50 s. If its length is increased by 50%, what will be its new period?3 marks
  5. A car's suspension can be modelled as a 350 kg mass supported by a spring. When a person of mass 70 kg gets into the car, the suspension compresses by 2.0 cm. Assuming the damping is negligible, what is the natural frequency of oscillation of the car on its spring?5 marks
  6. The displacement of an object in SHM is given by x = 0.040sin(50t), with x in meters and t in seconds. What is the first time after t=0 that the object's velocity is zero?3 marks
  7. An object in SHM has a total energy of 5.0 J and an amplitude of 12 cm. What is the kinetic energy of the object when its displacement is 6.0 cm?4 marks
  8. Explain why the motion of a bouncing ball is periodic but not simple harmonic.3 marks
  9. Two oscillators, A and B, start at the same time. Oscillator A starts at its positive amplitude, and oscillator B starts at its equilibrium position moving in the positive direction. They have the same frequency. What is the phase difference in radians between A and B, and which oscillator leads?2 marks
  10. A mass attached to a spring oscillates horizontally on a frictionless surface with period T. If the mass is replaced by a new mass that is four times larger, and the amplitude is halved, what is the new period of oscillation?2 marks

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